ANSWER KEY– LO 1.3

1. LO 1.3: The student is able to apply mathematical methods to data from a real or simulated population to predict what will happen to the population in the future. SP 2.2:The student can applymathematical routines to quantities that describe natural phenomena. Explanation: There are many formulas that can be used to calculate the growth rate of a population but to make an accurate prediction of a population’s growth, biologists use the logistic growth model. Unlike the other formulas used to calculate the growth of a population, such as the exponential growth model where resources are assumed to be limitless, the logistic growth model creates a prediction of what a population will look like over time with a limited amount of resources. The logistic growth model is similar to the exponential growth model except the per capita rate of increase of a population declines as the carrying capacity is reached. To decrease the per capita rate of increase the equation rmaxN(exponential growth rate equation) is multiplied by (K—N)/K where N is the total population size, rmax is the per capita growth rate of a population, and K is the carrying capacity of a population. M.C. Question: Logistic growth classifies a period of growth in a population where the growth rate ___________ as population size ____________. A) decreases, decreases B) decreases, increases C) increases, increases D) increases, decreases Learning Log/FRQ-style Question: A population of birds in Winston Salem was 3,500 in the year 2014. It was determined that the maximum per capita growth rate of the birds was .8 and the city could sustain a population of no more than 4,500 of the birds. What will the predicted population be in the year 2018? How far away will the bird population be from reaching its carrying capacity?

2. ANSWER KEY– LO 1.3 Multiple Choice Question: Logistic growth classifies a period of growth in a population where the growth rate ___________ as population size ____________. decreases, decreases decreases, increases increases, increases increases, decreases Learning Log/FRQ-style Question: A population of birds in Winston Salem was 3,500 in the year 2014. It was determined that the intrinsic rate of increase of the birds was .8 and the city could sustain a population of no more than 4,500 of the birds. What will the predicted population be in the year 2018? How far away will the bird population be from reaching its carrying capacity? ANSWER: From the calculations below we can determine that in the year 2018, the predicted population of the bird species is 4498 birds. It can also be concluded that 4500-4498=2 meaning that the predicted population will be 2 birds away from the carrying capacity of the birds in the ecosystem. All of the mathematical calculations were made using the data given to us in the original problem and plugging in the correct values for K, N, and rmax.

3. LO 3.4Describe representations and models illustrating how genetic information is translated into polypeptides SP 1.2 Describe representations and models of natural or man-made phenomena. Explanation: Genetic information is translated into polypeptides through the process of translation. During this process, messenger RNA (mRNA) produced by transcription is decoded by a ribosome complex to produce a specific amino acid chain, or polypeptide, that will later fold into an active protein. A biological model that demonstrates this process is the formation of collagen, or scar tissue proteins, which are created through the activation of genes in fibroblasts (connective tissue cells). In this naturally occurring phenomena, the genes in the DNA of fibroblast nucleus are activated and begin transcribing mRNA. This messenger RNA is translated into the collagen polypeptides, which are then secreted by the cell. • M.C. Question –Where does the process of translation occur in bacterial cells? • Within the cell’s cytoplasm • On the ER membrane • Within the cell’s nucleus • Within the mitochondria • F.R. Question- How does the process of polypeptide translation play a role in the replication of viruses? Slide created by Anna Hundley, 6th pd.

4. ANSWER KEY- LO 3.4 • M.C. Question –Where does the process of translation occur in bacterial cells? • Within the cell’s cytoplasm • On the ER membrane • Within the cell’s nucleus • Within the mitochondria F.R. Question- How does the process of polypeptide translation play a role in the replication of viruses? Viral replication begins with the injection of viral DNA into the host cell’s genome. DNA is then transcribed by the host’s organelles into mRNA. The ribosomal RNA then translates the mRNA into polypeptides. These polypeptides then form the outer shell of the virus, so that further replication of the viral DNA can occur within the cell. Slide created by Anna Hundley, 6th pd.

5. LO 3.6: The student can predict how a change in a specific DNA or RNA sequence can result in changes in gene expression. • SP 6.4: The student can make claims and predictions about natural phenomena based on scientific theories and models. • Explanation: In DNA and RNA mutations occur within the coding sequences. This can have many different effects and the protein produced depending on the mutation. There are many different kinds of mutations that can occur. One is point mutations which is the chemical change in just one base pair of a gene. An example of this would be a mutation of a single base pair in the gene that codes for one of the polypeptides of hemoglobin, which leads to sickle-cell disease. There are base-pair substitutions, which is the replacement of one nucleotide and its partner with another pair of nucleotides. Some substitutions are called silent mutations because of the redundancy of the genetic code, there is no change in the protein. Most substitutions are missense mutations, which the changed amino acid sequence will still code for the protein. The other affect is the nonsense mutation, and it causes the translation to be ended prematurely; making a smaller polypeptide sequence and most times a nonfunctional protein. Then there are insertions and deletions which are the addition or loss of nucleotide pairs. These mutations are worse then substitutions. This is because the loss of one nucleotide pair shifts the reading frame from that point down the nucleotide sequence. This is called a frameshift mutation. Scientists are able to make claims about the outcome of a genetic mutation in a coding sequence of amino acids because we know what mRNA codes for thanks to the dictionary of the genetic code. • M.C. Question: Which of the following mutations would most likely have a harmful effect? • a single nucleotide insertion near the start of a sequence • A base-pair substitution • A nucleotide deletion in an intron • A nucleotide deletion near the end of a sequence FRQ: A mutation resulting in the change in a nucleotide base pairing can have a detrimental effect on the protein produced. Explain why a change in the base pair in the beginning of a sequence is worse then a change towards the end.

6. Answers • M.C. Question: Which of the following mutations would most likely have a harmful effect? • a single nucleotide insertion near the start of a sequence • A base-pair substitution • A nucleotide deletion in an intron • A nucleotide deletion near the end of a sequence FRQ: A nucleotide change at the beginning of a sequence can have horrible results. This is due to the fact that a change early on in a sequence changes everything down the sequence. For example, you could have a 5’GAA3’, but that G is changed to a U, this codes for STOP, so the sequence is halted before it can really code a protein. Another example would be deleting a nucleotide, this causes frameshift that messes the mRNA’s reading thus creating a faulty protein. A mutation occurring towards the end is better because there is less nucleotides to mess up.

7. LO 2.3: The student is able to predict how changes in free energy availability affect organisms, populations and ecosystems. SP 6.4: The student can make claims and predictions about natural phenomena based on scientific theories and models. Explanation: Comparing trophic levels is usually done by analyzing how much total energy is contained by the trophic level rather than the total number of organisms, as the number of organisms decreases greatly from primary producers to the top level consumers. The average amount of energy that is transferred from trophic is only 10%, with 90% of the energy being released by other means like in the form of heat, and is frequently illustrated with energy pyramids. It is important to note that the autotrophs only receive 1% of the total energy from the sun. With only 1 out of every 10 units of energy effectively being transferred to the next trophic level the populations would be unable to obtain sufficient amounts of energy for each to survive. So in order to maintain enough healthy organisms the consumer population sizes decrease greatly with each successive level, which in turn allows the lower trophic level’s population to maintain a more constant size. There are usually only a few tertiary consumers in an ecosystem, correlating with their relatively low amount of total energy. • Multiple Choice Question: Out of the following species, which of their trophic levels is most likely to contain the most available energy that can be transferred to the next level? • A) Barn Owl • B) Cricket • C) Kentucky Bluegrass • D) Deer Mouse Free Response Question: A) Using several examples in your response, predict approximately how much free energy would be found in each trophic level from photoautotrophs to tertiary consumers if the sun has 4,000,000 kCal of free energy available. B) What would be some possible effects on population sizes and the ecosystem if all of a sudden a tertiary predator was only able to receive 5 percent of the free energy from its prey instead of 10 percent? C) Discuss two possible reasons that prevent more energy from being transferred to the next trophic level.

8. ANSWER KEY: Multiple Choice Question: Out of the following species, which of their trophic levels is most likely to contain the most available energy that can be transferred to the next level? A) Barn Owl B) Cricket C) Kentucky Bluegrass – CORRECT ANSWER D) Deer Mouse Free Response Question: • If the sun has 4,000,000 kCalof free energy available then photoautotrophs like Kentucky Bluegrass would be able to receive about 40,000 kCal from it. Using the 10 percent rule, an organism like a cricket that consumes grass like that would be able to obtain around 4000 kCal from the grass. A deer mouse that consumes those crickets would receive 400 kCal and its predator the barn owl could only get up to around 40 kCal of free energy. • If a top level predator like a bald eagle suffered some sort of damage and could only receive 5 percent of the total free energy instead of 10 percent then one bald eagle would need to consume double the amount of organisms it currently does in order to get the same amount of energy. This would negatively affect the ecosystem by greatly reducing the secondary consumer's population size, which would allow the primary consumer's population to grow unchecked by many predators. • While an organism is being consumed, a lot of the energy present is often lost as heat to the environment around the organism. The rest is used up by the consumer's body processes while it's consuming it's prey.

9. LO 1.22: The student is able to use data from a real or simulated population(s), based on graphs or models of types of selection, to predict what will happen to the population in the future. SP 6.4: The student can make claims and predictions about natural phenomena based on scientific theories and models. Explanation: Selection is defined as the process in which environmental or genetic influences determine which types of organisms thrive better than others. There are different types of selection, such as natural and artificial selection. Natural selection happens over several generations as a result of an environmental influence such as a change in niche. Artificial selection occurs when scientists choose the traits they want to succeed and breed organisms so that that specific trait will be present. In the case of natural selection, the animals’ environment changes in a way that makes it necessary for the species to evolve. Stabilizing selection as shown in the first image. Individuals with an intermediate phenotype are preserved and individuals with either extreme, either white or dark fur, will be less likely to survive. This occurs when the environment is an intermediate color and the animals need to blend in to survive. The second type of selection is directional selection. This type of selection favors individuals with the phenotype at one extreme. In the case demonstrated in the image, dark colored mice are favored so they can live among dark rocks and conceal themselves. The third type is disruptive selection. This type favors individuals at each end of the spectrum, but not intermediate individuals. This could occur when the species lives in an environment with light and dark rocks, but no intermediate colors. By examining the environment of the species and graphs of populations, an individual should be able to predict which type of selection is going to occur to save the species from extinction. M.C. Question: Which of the following statements concerning natural selection is FALSE? Stabilizing Selection produces equal numbers of all the phenotypes for a balanced gene pool. B. Directional Selection favors one extreme of the phenotype spectrum. C. Heterozygote Advantage Selection is the reason that some individuals are protected against the worst symptoms of malaria or sickle cell disease. D. Directional Selection is the most common type of selection. E. Stabilizing Selection removes extreme phenotypes from the population. FRQ-Style Question: Explain the unusual phenomenon of Bouvet Island where a species of worm evolved so that there were light colored worms and dark colored worms, but no medium colored worms. Draw a graph to help your explanation.

10. Answer Key- LO 1.22 Which of the following statements concerning natural selection is FALSE? • Stabilizing Selection produces equal numbers of all the phenotypes for a balanced gene pool. B. Directional Selection favors one extreme of the phenotype spectrum. C. Heterozygote Advantage Selection is the reason that some individuals are protected against the worst symptoms of malaria or sickle cell disease. D. Directional Selection is the most common type of selection. E. Stabilizing Selection removes extreme phenotypes from the population. Explain the unusual phenomenon of Bouvet Island where a species of worm evolved so that there were light colored worms and dark colored worms, but no medium colored worms. Draw a graph to help your explanation. The worms on Bouvet island have evolved due to a change in environment. Their soil has probably changed, due to a change in oxygen concentration, water presence, or mineral presence. As a result, the species of worm has evolved so that it will continue to blend in to the soil so that it will be less likely that birds will find the worms and eat them. The soil composition is different in different parts of the island, with lighter soil near the coast and dark soil inland. The worms that live closer to the coast will have the light coloration and the worms inland will have the dark coloration. This phenomenon is an example of disruptive selection, where variants with both phenotypic extremes are favored over the intermediate counterpart.

11. LO 1.19 The student is able to create a phylogenetic tree of a simple cladogram that correctly represents evolutionary history and speciation from a provided data set. SP 1.1 The student can create representations and models of natural or man-made phenomena and systems in the domain. Explanation: Cladograms show patterns of characteristics. If the patterns are due to common ancestry (each of the species is a descendent of the previous one), then the cladogram depicts evolutionary history and can also be considered part of a phylogenic tree. Such cladograms list the characteristics that developed through speciation under the species that first acquired that trait, showing how the original species (the outgroup) evolved over time. Other cladograms that list species that aren’t necessarily descendants of one another simply list any characteristic that differentiates one species from the previous ones. Cladograms are read from left to right, with the left-most species being the outgroup/original species and the right-most species being the most advanced from an evolutionary standpoint. Each species on the cladogram possesses the characteristic listed directly below it and all the ones before it, but none of the ones following it. Data sets for cladograms simply show which organisms possess what traits. To create a cladogram based on such a data table, one must find the species with the least number of traits and put it on the far left, find the species with the greatest number of traits and put it on the far right, and then put the remaining species in order from least to most amount of characteristics. Check to be sure that each species possesses all the defined traits that any species before it possesses. Finally, draw the basic cladogram skeleton and label it with specie names. Underneath the species names, name the characteristic that is newly acquired by the species. For more complicated cladograms, it may be necessary to add a branch to an existing branch. • Multiple Choice: If the characteristics of speciation on a cladogram are sexual reproduction, legs, live birth, and bipedal (in order from left to right), then which of the following lists of species is in the correct order (left to right) for this cladogram? • Archaea, lizards, fish, cats, humans • Archaea, fish, lizards, cats, humans • Humans, cats, lizards, fish, Archaea • Fish, lizards, Archaea, cats, humans • None of these Free Response: a)Draw the simplest and most logical cladogram for the species in the table below based on the data provided. b) Each of these species is part of a branch on a particular phylogenic tree. Which species is the most advanced and which is the least advanced from an evolutionary standpoint? Explain. c) If characteristics 1-4 are eyes, legs, fur, and no tail, respectively, then name possibilities for each of the species A-E.

12. Answer Key to LO 1.19 A C E B D 4 3 2 1 Characteristic Free Response: a)Draw the simplest and most logical cladogram for the species in the table below based on the data provided. b) Each of these species is part of a branch on a particular phylogenic tree. Which species is the most advanced and which is the least advanced from an evolutionary standpoint? Explain. c) If characteristics 1-4 are eyes, legs, fur, and no tail, respectively, then name possibilities for each of the species A-E. b) Species A is the least advanced because it doesn’t possess any of the characteristics on the cladogram. Species D is the most advanced because it possess all the characteristics on the cladogram. This means that Species D is the result of more evolutionary changes. c) (Many possibilities) Ex: Species A = jelly fish; Species C = dolphins; Species E = chameleons; Species B = rabbits; Species D = gorillas Multiple Choice: If the characteristics of speciation on a cladogram are sexual reproduction, legs, live birth, and bipedal (in order from left to right), then which of the following lists of species is in the correct order (left to right) for this cladogram? • Archaea, lizards, fish, cats, humans • Archaea, fish, lizards, cats, humans • Humans, cats, lizards, fish, Archaea • Fish, lizards, Archaea, cats, humans • None of these

13. LO 2.25 The student can construct explanations based on scientific evidence that homeostatic mechanisms reflect continuity due to common ancestry and/or divergence due to adaptation in different environments.SP 6.2 The student can construct explanations of phenomena based on evidence produced through scientific practices. MC Question: Which of the following statements concerning homeostatic mechanism is false? A)Environmental factors have an impact on homeostatic mechanisms. B)Pathways are an important part of the homeostasis of an organism. C)Homeostatic mechanisms aren’t derived from ancestors. D)Homeostatic mechanisms are used in multiple ways at a time in an organism. Explanation: All organisms in the animal kingdom are connected by the continuity of their homeostatic mechanisms or they can simply be divergent due to their homeostatic mechanisms. A homeostatic mechanism is a process that any living organism has to endure to keep a “balance” of all things occurring in the internal environment. An example could be perspiration while running. As you run your body is heating up, and then you perspire to cool yourself back down. These mechanisms are either passed down by common ancestors, or become diverged or experience a change when an adaption to a new environment comes into play. Natural selection would definitely take place when concerning homeostasis. A great example of how homeostasis diverges between organisms is the osmoregulatory process differences between freshwater and saltwater fish. Saltwater fish have a salt concentration in the water surrounding them that is much higher than the concentration that is needed for proper bodily function. The water itself is very dehydrating due to the salinity. So in order to release some salt they excrete out of their gills and urea extremely high levels of salt ions. On the other hand, you have freshwater fish who live in a very salt deficient environment but there is plenty of H2O. So in order to get the salt concentration that is needed for homeostasis they bring in salt ions from their gills and expel large amounts of salt free water by urea. This is a prime example of divergence of homeostatic mechanism divergence via environmental factors. But ways that mechanism are kept the same and passed on over time are by living in similar environments and the mechanism is successful and that trait is just passed on genetically. An example of an obvious one is the use of our lungs. When our blood lacks oxygen, we use our lungs to breathe in air and absorb oxygen. Comparisons can be made of different organisms homeostatic mechanisms by doing experiments and explaining any differences present by analyzing the evidence. Learning Log/FRQ style question: Suppose the bottlenose effect take place on a species of owl and the oxygen level drastically lowers in the air. Talk about the homeostatic mechanism used to get oxygen into the blood and then explain in detail the process that would have to take place for the population to have a resurgence after the crash following the environmental disaster.

14. Which of the following statements concerning homeostatic mechanism is false?A)Environmental factors have an impact on homeostatic mechanisms.B)Pathways are an important part of the homeostasis of an organism.C)Homeostatic mechanisms aren’t derived from ancestors.D)Homeostatic mechanisms are used in multiple ways at a time in an organism. Suppose the bottlenose effect take place on a species of owl and the oxygen level drastically lowers in the air. Talk about the homeostatic mechanism used to get oxygen into the blood and then explain in detail the process that would have to take place for the population to have a resurgence after the crash following the environmental disaster. When oxygen levels are depleated in the blood than the body must bring it back in using the lungs in the respiratory system. This gives the body homeostasis in terms of oxidation to keep the organs going and blood high in oxygen. What would have to happen for the owls to rise back in population is natural selection will have to take place. All of the owls who don’t have the lungs strong enough to produce enough oxygen will die out but then the ones that do will be able to pass on their genes and then all of the owls will have enherited a strong enough homeostatic mechanism for oxygen supply to the blood.

15. LO 3.15: The student is able to explain deviations from Mendel’s model of the inheritance of traits SP 6.5: The student can evaluate alternative scientific explanations. Explanation: Mendel’s model of inheritance included two alleles, one dominant and one recessive. Although this is true for a number of genes, many genes are inherited in other manners. Incomplete dominance “blends” the two allelic phenotypes together in a heterozygote (red flower + white flower = pink flower). In a heterozygote, codominance will display both phenotypes, such as a black and white cow. Sex-linked genes will not be present in a Mendelian ratio because males will have the recessive genotype/phenotype because they are hemizygous. Chloroplasts and mitochondria have their own DNA. Mitochondrial DNA is passed down through the ova. Therefore, all genes in the offsprings’ mitochondria will be inherited directly from the mother. M.C. Question: Someone with type AB blood will be demonstrating which type of inheritance? A.) Co-dominance B.) Mendelian inheritance C.) Incomplete dominance D.) Sex-linked Learning Log/FRQ-style Question: Red-green colorblindness is a sex-linked trait. If a mother is a carrier for the trait while a father is not affected, what are the chances of their daughter being colorblind? Their son? How likely is it that their daughter would be a carrier? Draw a punnett square to explain your answer. Incomplete Dominance

16. Answer Key- LO 3.15 Someone with type AB blood will be demonstrating which type of inheritance? A.) Co-dominance B.) Mendelian inheritance C.) Incomplete dominance D.) Sex-linked Red-green colorblindness is a sex-linked trait. If a mother is a carrier for the trait while a father is not affected, what are the chances of their daughter being colorblind? Their son? How likely is it that their daughter would be a carrier? Why is it more likely that their son would be colorblind? Draw a punnett square to explain your answer. The likelyhood of this couple’s son being colorblind is ½. Their daughter has no chance of being colorblind. However, she has a 50% chance of being a carrier and passing the trait down to her children. Males have a higher chance of having sex-linked disorders because they only have one X-chromosome. This means that they only need to get one recessive allele in order to have the disorder when women require two recessives in order to have the disorder. If the daughter is heterozygous for the disorder, then the chromosome containing the red-green colorblindness allele will become a Barr body and will not affect the daughter. Since the son only has one X-chromosome, one cannot become a Barr body and shut down if it is defective.

17. A. LO 3.24: The student is able to predict how a change in genotype, when expressed as a phenotype, provides a variation that can be subject to natural selection. B. SP 6.2: The student can make claims and predictions about natural phenomena based on scientific theories and models. SP 7.2: The student can connect concepts in and across domain(s) to generalize or extrapolate in and/or across enduring understandings and/or big ideas. C. ExplanationAn organism receives half of its genes from each parent, while genes are being transcribed there is possibility of incorrect transcription and so a mutation has occurred. These mutations are usually survivable but the organism may exhibit a change in phenotype. If the change in phenotype is advantageous to the type of environment the organism is in then the organism will survive and reproduce. This survival and reproduction allows for the possibility of the mutation to be passed on because it has been selected for by the environment. If the phenotype is selected against by the environment then the organism will not survive and reproduce and the mutation will not be passed on to future generations. For example, if a mutation occurs giving a bird a larger beak, the bird will be able to consume food more easily. This enables to bird for better survival and it will be able to reproduce and possibly pass on the mutation, allowing its descendants to be better equipped in their environment. D. Multiple Choice Question There is a specific mutation that causes a snake’s jaw to be unable to unhinge and therefore cannot swallow large food. What will mostly likely happen to the snake? a. The snake will survive and reproduce b. The snake will eat small prey c. The snake will not survive to reproduce d. The snake will become prey for other snakes E. Free Response Question Hummingbirds have a long, narrow beak that enables them to drink nectar from flowers. Explain how this adaptation was selected for over a long period of time, including the genetic basis for the change. http://dragonica.wikia.com/wiki/SI_Biology_What_happens%3F During meiosis a gene is considered a mutant if the nucleotide sequence of a parent was not copied correctly when passed to a gamete for a child. This diagram shows the incorrect pairing of a nucleotide. This nucleotide may cause a change in phenotype that will alter the organism’s chance of survival, whether it is helpful and allows the animal to survive, or harmful and causes the individual to lose fitness and become unable to survive and reproduce.

18. Learning Objective 3.24 Answer Key D. Multiple Choice Question There is a specific mutation that causes a snake’s jaw to be unable to unhinge and therefore cannot swallow large food. What will mostly likely happen to the snake? a. The snake will survive and reproduce b. The snake will eat small prey c. The snake will not survive to reproduce d. The snake will become prey for other snakes Due to the gap in the snake’s diet, it will not survive and therefore cannot reproduce. This eliminates the mutation from the gene pool because it was selected against by the environment. E. Free Response Question Hummingbirds have a long, narrow beak that enables them to drink nectar from flowers. Explain how this adaptation was selected for over a long period of time, including the genetic basis for the change. There must have been an error in the copying of DNA during meiosis that led to a mutation in the DNA sequence. This would have caused the phenotype of the beak to be slightly longer than normal. When this individual fed on the flower nectar it was able to reach the food source more easily and therefore could survive to reproduce. This means the trait was selected for by the environment of the hummingbirds because it allowed the individual to have an adaptation that made its survival easier than it would be if it had not had the mutation for a longer beak. Because this individual survived and could reproduce, its offspring have the possibility of inheriting the mutation. When the mutation is passed on it helps all those who have it to survive over those that do not. This increases its frequency in the hummingbird population and over a long period of time, increases the average length of the overall hummingbird species.

19. LO 1.5: The student is able to connect evolutionary changes in a population over time to a change in the environment. SP 7.1: The student can connect phenomena and models across spatial and temporal scales. Explanation: In insects, for example, natural selection occurred when the insects were exposed to a insecticide. The insects that did not have the ability to resist the insecticide, were killed. Those who actually survived the insecticide, passed on their genetic traits that allowed them to survive to their offspring. In some cases, an insect is resistant because their enzymes can break down the toxic chemicals into less toxic chemicals, allowing the insect to survive. In other cases, resistance is due to a mutation on the insect’s autosomes. Over time, the insect’s population will slowly all become resistant (because those who are not will die) and the insecticide in the environment will have no effect on them. M.C. Question: Which statement best explains evolution? A. Survival of the fittest B. When a species reaches equilibrium C. A change in inherited characteristics over time D. All the genes in a species stay the same Learning Log/FRQ-style Question: Suppose a farm that has had insect problems buys a new insecticide. Explain what would happen to the insects over time and then draw a graph, showing the insect’s population before and after the insecticide had been introduced.

20. Answer Key LO 1.5 M.C. Question: Which statement best explains evolution? A. Those who survive are the fittest and produce more offspring B. When a species reaches equilibrium C. A change in inherited characteristics over time D. All the genes in a species stay the same Learning Log/FRQ-style Question: Suppose a farm that has had insect problems buys a new insecticide. Explain what would happen to the insects over time and then draw a graph, showing the insect’s population’s numbers over time. At first exposure, the insects without any resistance already will die. Those who have the alleles that will allow them to survive, will survive. Those who do survive will mate with each other and the allele that had originally allowing them to survive against the insecticide, will be passed on to their offspring and their offspring will be resistant as well.Also, their offspring will pass it on to theirs, the 3rd generation, meaning, the species will pretty much by then be totally resistant and the insecticide will be harmless to the insects and not even work anymore.

21. Learning Objective 2.35 • A) The student is able to design a plan for collecting data to support the scientific claim that the timing and coordination of physiological events involve regulation. • B) Science Practice 4.2: The student can design a plan for collecting data to answer a particular scientific question. • C) The student should be able to create an experiment to collect the data needed to support the claim that physiological events sensitive to timing and coordination (i.e. metabolism, mitosis, osmosis, action potential, transpiration, flowering…) involve regulation. • D) You buy a chrysanthemum you want to plant in the back yard. You know it is a short-day plant, so it needs longer dark night hours than day hours to flower. With this in mind, you plant it in the winter and you live in region with very mild winters. However, the plant is not flowering. What is the most likely reason for this? • A- The day’s sun is so bright, that the chrysanthemum thinks the day is longer than it is. • B- Your son often sets off fireworks in the backyard, disrupting the chrysanthemum’s critical dark period. • C- There’s a genetic defect in the plant increasing transpiration at night that causes the plant’s cells to become flaccid, so the flower petals become limp and can’t open. • D- Your neighbor’s metal rock band practices in his garage late at night often and the sound vibrations disrupt the chrysanthemum’s flowering cycle, preventing it’s blooming. • E) (a) Your AP Biology teacher asks you to brainstorm possible experiment ideas that would prevent cells from completing the cell cycle. Explain a hypothesis to test. (b) Cyclin regulates the G1 and S phase of the cell cycle. Explain cyclin’srole in the cell cycle, and how dysfunction could cause the abnormal growth of tumors.

22. Diagrams

23. Answer Key • You buy a chrysanthemum you want to plant in the back yard. You know it is a short-day plant, so it needs longer dark night hours than day hours to flower. With this in mind, you plant it in the winter and you live in region with very mild winters. However, the plant is not flowering. What is the most likely reason for this? • A- The day’s sun is so bright, that the chrysanthemum thinks the day is longer than it is. • B- Your son often sets off fireworks in the backyard, disrupting the chrysanthemum’s critical dark period. • C- There’s a genetic defect in the plant increasing transpiration at night that causes the plant’s cells to become flaccid, so the flower petals become limp and can’t open. • D- Your neighbor’s metal rock band practices in his garage late at night often and the sound vibrations disrupt the chrysanthemum’s flowering cycle, preventing it’s blooming. (a) Your AP Biology teacher asks you to brainstorm possible experiment ideas that would prevent cells from completing the cell cycle. Explain a hypothesis to test. (b) Cyclin regulates the G1 and S phase of the cell cycle. Explain cyclin’s role in the cell cycle, and how dysfunction could cause the abnormal growth of tumors. • (a) If the transcription of cyclin is prevented (either from a mutation or blocking transcription) or an enzyme degrades the cyclin protein too early, then it cannot bond to Cdk to produce MPF’s, which signal a cell to begin mitosis. • (b) When cyclin binds with Cdk to form MPF’s, the cell passes the G2 check point in the cell cycle and continues on to Mitosis. Then, the cyclin is degraded and the cell enters G1 then Interphase. If the cyclin is not denatured, then mitosis and continued replication of those cells will continue causing abnormal growth of tumors.

24. LO 2.22 The student is able to refine scientific models and questions about the effect of complex biotic and abiotic interactions on all biological systems, from cells ad organism populations, communities and ecosystems.SP 1.3 The student can refine representations and models of natural or man-made phenomena and systems in the domain.SP 3.2 The student can refine scientific questions. Explanation: The global distribution of organisms broadly reflects the influence of abiotic factors such as regional differences in temperature, water, and sunlight. Environmental temperature, for example, plays a role in the cells’ ability to work properly because it can easily rupture or denature if it is not set at a specific temp. Along with that, an organism’s internal temperature is affected by heat exchange with its environment. Most organisms need to maintain their internal temperature in order for their bodies to function best. Ability for osmoregulation also factor in because global distribution reflects their ability to obtain and conserve water. Sunlight is essential because light effects the development and behavior of the organisms that are sensitive to photoperiod. Most importantly, sunlight provides the energy that drives nearly all ecosystems because of autotrophs and other photosynthetic organisms that use this energy source directly. Since autotrophs and photosynthetic organisms are the ultimate sources of organic compounds for all non autotrophic organisms, they are referred to as the producers of the biosphere. These producers are critical in the stability of populations, communities, and ecosystems. The structure and dynamics of a community depend largely upon the transfer of food energy up the trophic levels from these primary producers. We can summarize the trophic levels of a community by diagramming a food web. M.C. Question: Which question most relates to how biotic and abiotic interactions in an ecosystem? a. How does sunlight interact with autotrophs? b. How does temperature affect wind patterns? c. How does prey and predator effect the food web? d. How does sunlight intensity predict weather change? FRQ: The diagram provided displays a food web. How would the ecosystem be affected if the butterfly orchid population were to drastically decline? How would the abiotic environment be affected?

25. Answer Key LO 2.22 M.C. Question: Which question most relates to how biotic and abiotic interactions in an ecosystem? a. How does sunlight interact with autotrophs? b. How does temperature affect wind patterns? c. How does prey and predator effect the food web? d. How does sunlight intensity predict weather change? FRQ: The diagram provided displays a food web. How would the ecosystem be affected if the butterfly orchid population were to drastically decline? How would the abiotic environment be affected? The mosquito population would decrease because they rely on the Butterfly Orchids as a food source. Since the population of the mosquitos declined, the Southern Leopard Frog population would decrease as food sources become scarce. The Grass carp would have to rely on Bladderwort as a food source, due to the absence of mosquitos, therefore decreasing the Bladderwort population. The soil would be affected since there are less butterfly orchids consuming the nutrients, allowing different plant species to thrive.

26. LO 3.7: The student can make predictions about natural phenomena occurring during the cell cycle. SP 6.4: The student can make claims and predictions about natural phenomena based on scientific theories and models. Explanation: The cell cycle is a predictable pattern undergone by most eukaryotic cells. If you know the progression of the cell cycle, you can predict, for example, what phase a cell is in by knowing whether MPF levels are increasing or decreasing. The cell cycle consists of five phases: G1, S, G2, M, and C. During G1, the cell grows. At some point during G1, the G1 checkpoint occurs to determine whether the cell will divide, delay division, or enter a resting stage. If the cell decides to divide, it enters the S (replication) phase. It copies all of its DNA so that it has two copies of each chromosome. Then cells go into the G2 phase, where they grow and prepare for division. During G2, cells gradually accumulate G2 cyclin. The cell reaches the G2 checkpoint and checks whether the DNA is successfully replicated. At this checkpoint, Cyclin-dependent kinase binds with the G2 cyclin to form Mitosis Promoting Factor (MPF). MPF phosphorolates proteins that cause the cell to end G2 and begin the M (mitosis) phase. MPF also activates proteins that destroy G2 cyclin, so MPF levels drop during the M phase, causing the end of mitosis. In mitosis, the cell produces two daughter cells, each with a complete set of chromosomes. The M checkpoint assesses the progress of mitosis. At the end of the M phase, the cell divides into the two daughter cells. M.C. Question: A researcher is making several measurements on a particular mouse liver cell. He finds that the MPF concentration is steadily rising, and that there is 2X as much DNA in the cell nucleus as he measured earlier. Based on this information, what stage of the cell cycle is this liver cell most likely undergoing? a) mitosis (M) b) DNA synthesis (S) c) G1 d) G2 FRQ-Question: Suppose a mutation occurs in a pancreas cell that causes G2 cyclin to be misshapen and unable to bind to CDK. Describe the immediate effects (at the molecular/cell level) of this mutation on the normal cell cycle of this pancreas cell.

27. Answer Key (LO 3.7) M.C. Question: A researcher is making several measurements on a particular mouse liver cell. He finds that the MPF concentration is steadily rising, and that there is 2X as much DNA in the cell nucleus as he measured earlier. Based on this information, what stage of the cell cycle is this liver cell most likely undergoing? a) mitosis (M) b) DNA synthesis (S) c) G1 d) G2 FRQ-Question: Suppose a mutation occurs in a pancreas cell that causes G2 cyclin to be misshapen and unable to bind to CDK. Describe the immediate effects (at the molecular/cell level) of this mutation on the normal cell cycle of this pancreas cell. An inability of cyclin to bind to CDK would prevent the formation of mitosis promoting factor (MPF). An inability to produce MPF , and therefore an inability to reach the MPF threshold, would in turn prevent the phosphorolyzation of proteins involved in mitosis, making mitosis impossible. The cell might continue growth until lysis, or enter a resting stage, or perhaps undergo apoptosis.

28. SP 5.3 The student can evaluate the evidence provided by data sets in relation to a particular scientific question. (Example) LO 1.18 The student is able to evaluate evidence provided by a data set in conjunction with a phylogenetic tree or a simple cladogram to determine evolutionary history and speciation. Explain: Cladograms and phylogenetic trees are visual displays of hypothesized routes in which multiple species have evolved from. The species at the left end is the “starting” species and the other on the right is the more evolved species. This is usually determined by the number of amino acid differences between the organisms being compared. The less amount of amino acid differences between organisms the more similar their DNA is to each other, therefor the closer those species are related. Cladograms can also be separated by homologous attributes that the organism’s possess. Each high tiered organism possesses a evolutionary attribute from the previous. Speciation which occurs during the evolutionary process is noted by an ascending line off the species that it evolved (example D and E) M.C. Question Scientist are researching a specific medication that targets proteins this will be used on humans but first they want to test the medication on animals and see how they respond, use table 1 in order to determine which animal would have the most similar proteins for accurate testing. A) Chimpanzee B) Gorilla C) Mouse D Monkey E) Rabbit FRQ Draw a cladogram to represent the pathway of evolution from the Pictures below, in addition separate them with their evolutionary Features that differentiate them. (% similarity to human DNA) use for M.C question Table 1

29. Answer Key LO 1.18 • M.C. : The correct answer is A because the percentage of DNA similarity to humans is the highest meaning the the chimp is most likely to have similar matching sequence. • FRQ: The response must look similar to the expected answer below and must be separated by the physical features.

30. LO 3.8The student can describe the events that occur in the cell cycle. SP 1.2 The student can describe representations and models of natural or man-made phenomena and systems in the domain. Explanation The cell cycle is the process by which eukaryotic cells reproduce and pass on their genetic information. The cell cycle has 2 parts: interphase and mitosis. Interphase occurs in 3 phases: G1, S, and G2. The first growth phase, or G1 phase, is the first part of interphase in which DNA synthesis has not yet begun. The synthesis phase, or S phase, is the part of the cell cycle in which DNA is replicated. The second growth phase, or G2 phase, is the part of interphase that occurs after DNA synthesis. A quick summary of Interphase: cell grows (G1), copies chromosomes (replicates DNA) while continuing to grow (S), and keeps growing (G2) as well as preparing for the cell division that occurs in Mitosis. In Mitosis, the cell goes through 5 different phases: prophase, metaphase, anaphase, telophase, and cytokinesis. In prophase, the nuclear membrane as well as the nucleolus breakdown and the DNA form chromosomes. These chromosomes have kinetochores (near the center) where the microtubules of the mitotic spindle are able to attach themselves. This is crucial, for in metaphase the spindle microtubules move the chromosomes and align them in the center of the cell. During anaphase, the microtubules shorten, separating the chromosomes into chromatids and drawing them to opposite ends of the cell. During telophase, the chromatids continue on until they reach the poles of the cell, all the while the cell is elongating. Cytokinesis occurs when (in animal cells) a cleavage furrow forms in the center of the parent cell, and then divides it into two genetically identical daughter cells. M.C. Question Which of the following correctly lists the order of the phases of the cell cycle? A) G1, G2, S, M B) M, G1, G2, S C) G1, S, G2, M D) S, G1, G2, M FRQ Question a) Explain the importance of the cell cycle in eukaryotic cells. b) List, in order, the phases of Mitosis. Explain each phase thoroughly. You may draw a diagram to help explain each phase.

31. Answer Key - LO 3.8 M.C. Question Which of the following correctly lists the order of the phases of the cell cycle? A) G1, G2, S, M B) M, G1, G2, S C) G1, S, G2, M → correct answer D) S, G1, G2, M (Sample Diagram. See FRQ Part b.) FRQ Question a) Explain the importance of the cell cycle in eukaryotic cells. b) List, in order, the phases of mitosis. Explain each phase thoroughly. You may draw a diagram to help explain each phase. Correct Response a) In eukaryotic cells, the cell cycle is what enables cells to reproduce new genetically identical cells as well as replace and repair old, damaged cells. b) The phases of mitosis are as follows: prophase, metaphase, anaphase, telophase, and cytokinesis. During prophase, the nuclear membrane and nucleolus break down, while the DNA unravels and becomes chromosomes. During metaphase, the spindle microtubules connect themselves with the kinectochores located on the chromosomes, and align the chormosomes in the center of the cell. In anaphase, the spindle microtubules begin to shorten with the help of motor proteins, and the chromosomes are saparated. These sister chromatids are each drawn towards opposite ends of the cell. In telophase, the chromatids continue on towards opposing sides of the cell, and a cleavage furrow begins to form. In cytokinesis, the cleavage furrow saparates, leaving two genetically identical daughter cells.

32. LO 3.40- The student is able to analyze the data that indicate how organisms exchange information in response to internal changes and external cues, and which can change behavior. SP 5.1- The student can analyze data to identify patterns or relationships. Explanation- Over time, organisms have developed the ability to sense and respond to external cues from other organisms or their environment. These adaptations can increase the likelihood of survival and reproduction. They also affect the nervous system to produce responses or changes in behavior, such as the rate of metabolism or respiration, according to their stimuli. Other examples of behaviors dependent on outside conditions include biological rhythms, mating behaviors, and animal communication. For these changes to occur, the nervous system takes processed information from the brain and sends signals to other parts of the body. The behavior of an individual affects the behavior of a population to ensure survival of the group. For example, when a meerkat notices a predator approaching their habitat, their brain processes a threat and the sympathetic division of the autonomic nervous system increases heart rate, raises blood pressure, and dilates pupils due to the stress and fear. The meerkat then responds by warning the rest of the population with a call so the others know to hide and can avoid being taken by the predator. Once the threat passes, the parasympathetic system decreases heart rate and blood pressure to calm them down and maintain homeostasis. Multiple Choice: Which of the following is not a population behavior caused by an individual’s response to external cues? Meerkats hiding when hearing a warning call by the first meerkat that sees a predator Bee stinging an intruder of the nest to protect the rest of the colony and the queen Pack of lions following one lion that has caught a scent of a wildebeest to kill and eat School of young beta fish darting in the opposite direction when seeing the male beta fish vibrating its fins to signal danger Free Response: Interpret the following graph between a period of time and the heart rate of an organism at the sight of a predator. Describe the events of the autonomic nervous system occurring at each of the labeled points of the graph. Propose a hypothetical situation of the processes that are occurring. (Be specific.) Time vs. Heart Rate B C Heart Rate D A Time (minutes)

33. Answer Key- L.O 3.40 • Multiple Choice: • Which of the following is not a population behavior caused by an individual’s response to external cues? • Meerkats hiding when hearing a warning call by the first meerkat that sees a predator • Bee stinging an intruder of the nest to protect the rest of the colony and the queen • Pack of lions following one lion that has caught a scent of a wildebeest to kill and eat • Young beta fish darting in the opposite direction when seeing the male beta fish vibrating its fins to signal danger Time vs. Heart Rate B Heart Rate C • Free Response: • Interpret the following graph between a period of time and the heart rate of an organism at the sight of a predator. Describe the events of the autonomic nervous system occurring at each of the labeled points of the graph. Propose a hypothetical situation of the processes that are occurring. (Be specific.) D A Time (minutes) At point A, the animal is performing its normal activities and heart rate is normal. During point B, the animal notices the predator and processes it as a danger, so its sympathetic nervous system arouses, increasing heart rate drastically and rapidly. The organism responds accordingly to the stimuli by hiding, fighting, or running. By point C, the predator has passed so the parasympathetic system slows down the heart rate back to normal to maintain homeostasis. At point D, heart rate has returned to normal and the animal returns to their daily activities. A hypothetical situation of this process may be an antelope spotting a lion. As the antelope is eating grass, it sees a lion ready to attack and the sympathetic nervous system kicks in to signal a response such as running. Once the lion is tired out and leaves the antelope alone, the antelope can then rest and allow its heart rate to return to normal via the parasympathetic nervous system. It then continues to eat grass because the threat or stimuli is gone and the antelope has recovered.

34. LO 4.2: The student is able to refine representations and models to explain how the subcomponents of a biological polymer and their sequence determine the properties of that polymer. SP 1.3: The student can refine representations and models of natural and man-made phenomena and systems in the domain. Explanation: DNA molecules, for example, are composed of many connecting nucleotide bases to form a double helix of genetic information. This information is coded in sets of nucleotides and is “decoded” by transcription and translation. The nucleotide bases are not capable of carrying genetic information on its own, however, when they come together, they become one big natural polymer to code for different genes in the DNA. Other polymers are formed in the same manner; they are composed of linked monomers to create a larger polymer. Polymer meaning many monomers. DNA gyrase cuts the double helix open and the strands separate. Helicase then comes and unzips the double stranded DNA. Binding proteins temporarily bind to each strand to keep them separated. After, DNA polymerase comes along and “walks” down the strands and adds new complementary nucleotides. A subunit of the polymerase then proofreads the new strands and DNA ligase seals the fragments. The new DNA winds up again into its double helix Shape. Another polymer, glycogen, is made of many monomers of glucose, a simple sugar. When your blood glucose levels go down, your body looks to break down glycogen into glucose so your blood sugar can rise to homeostasis. However, if your blood sugar rises to levels too high, then your brain triggers your pancreas to produce insulin and lower your blood sugar. Multiple Choice Question: Which of these statements is true about DNA and its components? A. It codes for genes in sets of three nucleotide bases B. DNA is shaped like a flat staircase C.DNA carries phenotypic information only D.DNA carries genotypic information only E. All statements are true. FRQ- Style Essay Question: Name two polymers and how its subcomponents give it its properties.

35. Answer Key – LO 4.2 Which of these statements is true about DNA and its components? A. It codes for genes in sets of three nucleotide bases B. DNA is shaped like a flat staircase C.DNA carries phenotypic information only D.DNA carries genotypic information only E. All statements are true. FRQ-Styled Question: Name two polymers and how its subcomponents give it its properties. One polymer is glycogen which is composed of many monomers of glucose. This structure gives it the properties to raise blood sugar when broken down into glucose again. The other polymer is polypropylene, made out of many monomers of polypropylene. It is used as a plastic in many plastic container, tupperware, and rugs because of it's high melting point. Meaning if you put hot soup in a tupperware of this kind of plastic, it wont warp.

36. LO 2.6: The student is able to use calculated surface area-to-volume rations to predict which cell(s) might eliminate wastes or produce nutrients faster by diffusion. SP 2.2: The student can apply mathematical routines to quantities that describe natural phenomena. Explanation: In math, the formula for finding the surface area of a sphere, or cell, is known as A=4∏r^2; r being the radius; however for ∏ we will be using 3.14. The volume of a sphere is known by the formula V=⁴⁄₃∏rᶟ. In a eukaryotic/prokaryotic cell, the surface area can determine how much waste can be released at one time through the cell membrane and the volume can be used to determine exactly how much waste is released in relation to volume before diffusion and after to get the difference. The volume-to-surface area ratio of a cell can determine the amount released because the smaller the cell, the more evenly distributed the cell can get food and release waste. However, if a cell is bigger, the surface area: volume (ratio) becomes more uneven since the volume of a cell can exceed its surface area which may result in the cell dying due to there not being enough surface area to keep up with the increasing volume. M.C.: The surface area of an animal cell is 200.96. what is the radius? What is the volume? Round to the nearest hundredth. 4; 268 6; 267.95 4; 267.95 6; 904.32 Short FRQ: A cell is observed and determined to be 2.5cmᶟ. What is the surface area and volume of the cell to the nearest tenth? Would more waste/food be released/absorbed into the cell if it were bigger in size or smaller? Explain. (Think about the surface area-to-volume ratio of a cell).

37. ANSWER KEY- LO 2.6 The surface area of an animal cell is 200.96. what is the radius? What is the volume? Round to the nearest hundredth. 4; 268 6; 267.95 4; 267.95 6; 904.32 A cell is observed and determined to be 2.5cmᶟ. What is the surface area and volume of the cell to the nearest tenth? Would more waste/food be released/absorbed into the cell if it were bigger in size or smaller? Explain. (Think about the surface area-to-volume ratio of a cell). The surface area of that cell would be 78.5cmᶟ and the volume would be 65.42cmᶟ. More waste/ food would be released/absorbed if the cell was smaller because the cell would have a more equal ratio of waste/food being released/absorbed than if it was bigger and diffusion can happen at a more equal rate which is why our cells are so tiny. This also explains why cells grow to a certain size then go through diffusion rather than growing more in size.